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Is (-2, 6) a solution to the system of these linear equations: x + 2y = 10 and 3x + y = 0? Why? (1 point) No, because the graphs don't intersect at (-2, 6). Yes, because the graphs intersect at (-2, 6). Yes, because the graphs don't intersect at (-2, 6). No, because the graphs intersect at (-2, 6).
Yes, because the graphs intersect at (-2, 6).
To determine if (-2, 6) is a solution to the system of linear equations x + 2y = 10 and 3x + y = 0, we can substitute the values of x and y into both equations and check if the equations are satisfied. Let's do that:
For the first equation, x + 2y = 10, substituting x = -2 and y = 6 gives:
(-2) + 2(6) = 10
-2 + 12 = 10
10 = 10
The equation is satisfied.
For the second equation, 3x + y = 0, substituting x = -2 and y = 6 gives:
3(-2) + 6 = 0
-6 + 6 = 0
0 = 0
The equation is also satisfied.
Therefore, (-2, 6) is a solution to the system of linear equations.
The answer is: Yes, because the equations are satisfied when we substitute (-2, 6) into them.
To determine if (-2, 6) is a solution to the system of linear equations, we need to substitute the values of x and y into both equations and check if the equations hold true.
The given system of equations is:
1) x + 2y = 10
2) 3x + y = 0
Let's substitute the values x = -2 and y = 6 into both equations:
1) Substitute x = -2 and y = 6 into equation 1:
-2 + 2(6) = 10
-2 + 12 = 10
10 = 10
The equation holds true.
2) Substitute x = -2 and y = 6 into equation 2:
3(-2) + 6 = 0
-6 + 6 = 0
0 = 0
The equation holds true.
Since (-2, 6) satisfies both equations, it is indeed a solution to the system of linear equations. Hence, the answer is "Yes, because the graphs intersect at (-2, 6)."